Re: Please, PLEASE, hold your questions/comments/elsewhat til the end. Thank you. :)

Threshold wrote:
) Willem wrote:
)
) Okay, here's the problem: One of the intermediates *will* inevitably
) have
) the same MD5 as the original file. Furthermore, on average it will
) happen
) before you have shaved off enough bits to match the size of that MD5
) hash.
)
) -
)
) I don't think so. I'm aware that it's a possibility, but given that
) there are more files of N length, and/or of N-1 length with a different
) MD5 than the same one, I think it's more probable that an algorithm is
) derivable such that no intermediates share an MD5 with the original.

First of all, you seen to be making several unproven asumptions.

But more importantly, there is another flaw in your scheme:

You propose an algorithm that takes a file of size 2^N, and for
half of the possible files it outputs a file of size 2^(N-1), and
for the other half it outputs a file of size 2^N, am I correct ?

I have a very easy algorithm that does that:

- If the last bit is a 1, make the last bit a 0.
- Otherwise, remove the last bit.

Disregarding the MD5 stuff, does this algorithm fit your description ?


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
.

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